DROPDOWN dynamically feed

I am using chronoforms v6 and noticed that in many cases there is no need to create extra code for some functions.
I am counting the number of records in a database and hovering LOOP to generate a list for a DROPDOWN ...
Example: I find that I have 4 records in a database, how to dynamically feed a DROPDOWN with the "1,2,3,4" choice options?
Is there a way to implement without generating extra code?

There's a dynamic dropdown example in the demo forms.

I am using a PHP to populate the dropdown with a count of database records, one same works well plus I have to get the maximum value of records that comes from the amount of records in the database. The value is in read_data15 it indicates that it is number 3, however when using in "for" this value is ZERO ...

​

//return array("Ponto 1","Ponto 2","Ponto 3","Ponto 4",);
$results = array();
$quant = '{var:read_data15}';
$results

= 'Selecione o ponto desejado';
for ($i = 1; $i <= $quant; $i++) {
$results

= 'Ponto '.$i;
}
return $results;

​

how can I get around this problem?

//return array("Ponto 1","Ponto 2","Ponto 3","Ponto 4",);
$results = array();
$quant = '{var:read_data15}';
$results

= 'Selecione o ponto desejado';
for ($i = 1; $i <= $quant; $i++) {
$results

= 'Ponto '.$i;
}
return $results;

var:read_data.model

return '{var:read_data15}';

​

by simply testing with this I get the value 3 ...

Healyhatman...

​

thus ?

​

$results = array();
$quant = '{var:read_data15.model}';
$results[ ] = 'Selecione o ponto desejado';
for ($i = 1; $i <= $quant; $i++) {
$results[ ] = 'Ponto '.$i;
}
return $results;

m are you sure it's 3, or is it an array containing the number 3

I am counting records through the option below in read_data15 and currently have 3 records ... I need this value inside the "for" to be able to create 3 elements in the array that goes to the dropdown when returning from PHP

​

the read_data15...

​

Select type = Return the count of records matching the filtering conditions.

Oh sorry I see the problem. You can't use shortcode a in PHP.

​

{var:} should be $this->get("variable name", "default if not found");

any problem to overcome the problem?

I just told you? What is it you're asking?

Thanks, follow a link to video with the implementation.

​

https://youtu.be/mkr_460kmuM

$quant = $this->get("read_data15", 0);

​

​

You didn't read what I wrote properly, $this->get takes two parameters - the name of the variable you want to retrieve, and a default value if the variable is not found.

Also try doing

​

return json_encode($results);

​

and in the dropdown value put

​

{var/jsonde:php17}

Thanks a lot for the guidance, you are correct ... however the implementation described in the video is working correctly ... at the time I noticed that $ this in the chronoforms is an access to the entire structure of the project simply by knowing how to access the desired variables, thank you...

Throughout my code I will certainly have other doubts, thank you very much.